Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

232so that we define the map φ F (a, b) : Q F A (a, b) =⋃a ′ ∈F ← (b)A (a, a ′ ) → BQ F (a, b) =B (F (a) , b) = B (F (a) , F (a ′ )). Clearly, such a map is induced by the matrix mapA (a, a ′ ) → B (F (a) , F (a ′ ))f ↦→ F (f) .Since F is a functor, F preserves composition, F (g ◦ f) = F (g) ◦ F (f) , i.e. F iscompatible with respect to the multiplications of the monads (A, M) and (B, N),and F preserves the identity, F (1 a ) = 1 F (a) , i. e. F is compatible with respect to theunits of the monads (A, M) and (B, N). Hence we get that F is a monad functor.Let now F, G : (A, M) → (B, N) be monad functors and let χ : (F, φ F ) → ( G, φ G)be a functor transformation. Then we have that χ : Q F → Q G is defined by setting{∅{(a, F (a))}Then, we have Q F (a, b) Q F (f)−→ Q F (a ′ , b ′ )χ (a, b) : Q F (a, b) → Q G (a, b)} {if b ≠ F (a)∅↦→if b = F (a) {(a, G (a))}}if b ≠ G (a).if b = G (a){∅{(a, F (a))}Q F (a, b) Q F (f)−→ Q F (a ′ , b ′ )} {}if b ≠ F (a)∅ if b↦→′ ≠ F (a ′ )if b = F (a) {(a ′ , F (a ′ ))} if b ′ = F (a ′ )and Q G (a, b) Q G(f)−→ Q G (a ′ , b ′ ) we have thati.e.χ is a monad functor transformation.Now, let us define the following mapχ (a ′ , b ′ ) (Q F (f)) = Q G (f) (χ (a, b))F (C) : Mnd (C) → BIM (C)(X, A) ↦→ (X, A)(X, A) (Q,φ)→ (Y, B) ↦→ (Q · A, (1 Q · m A ) (φ · 1 A ) , 1 Q · m A )(Q, φ) σ → (P, ψ) ↦→ σ · 1 A .Proposition 1<strong>1.</strong>15. The map F defined above is well-defined and it is a pseudofunctor.Proof. First, let us prove that (Q · A, (1 Q · m A ) (φ · 1 A ) , 1 Q · m A ) is a bimodule. Infact, we haveλ Q·A (1 B · λ Q·A ) = (1 Q · m A ) (φ · 1 A ) (1 B · 1 Q · m A ) (1 B · φ · 1 A )φ= (1 Q · m A ) (1 Q · 1 A · m A ) (φ · 1 A · 1 A ) (1 B · φ · 1 A )m A ass= (1 Q · m A ) (1 Q · m A · 1 A ) (φ · 1 A · 1 A ) (1 B · φ · 1 A )(233)= (1 Q · m A ) (φ · 1 A ) (m B · 1 Q · 1 A ) = λ Q·A (m B · 1 Q · 1 A )